Elements $I_n+pXinmathbb{Z}^{ntimes n}$ with finite order in $operatorname{SL}(n,mathbb Z)$
$begingroup$
Given $p$ an odd prime, $Xinmathbb{Z}^{ntimes n}$ a matrix.
If $I_{n}+pXin operatorname{SL}(n,mathbb Z)$ has finite order, prove $X=0$.
My idea:
let $m=p^rh$, $(p,h)=1$, s.t. $(I+pX)^m=I$.
let $(I+pX)^{p^r}=I+pY$, $I=(I+pX)^m=(I+pY)^h=(I+hpY+p^2YZ)$. $Y(hI+pZ)=0$.
If we can prove $Y=0$, then the following would be quite easy.
So how to prove $hI+pZ$ is invertible? Thanks for your time and help!
abstract-algebra group-theory
asked Nov 24 '18 at 13:06
AndrewsAndrews
3831317
$endgroup$
add a comment |
$begingroup$
Given $p$ an odd prime, $Xinmathbb{Z}^{ntimes n}$ a matrix.
If $I_{n}+pXin operatorname{SL}(n,mathbb Z)$ has finite order, prove $X=0$.
My idea:
let $m=p^rh$, $(p,h)=1$, s.t. $(I+pX)^m=I$.
let $(I+pX)^{p^r}=I+pY$, $I=(I+pX)^m=(I+pY)^h=(I+hpY+p^2YZ)$. $Y(hI+pZ)=0$.
If we can prove $Y=0$, then the following would be quite easy.
So how to prove $hI+pZ$ is invertible? Thanks for your time and help!
abstract-algebra group-theory
asked Nov 24 '18 at 13:06
AndrewsAndrews
3831317
$endgroup$
2
$begingroup$
Have you already done anything? Did you try some examples, or some cases of $;p;$ ? Anything at all?
$endgroup$
– DonAntonio
Nov 24 '18 at 13:12
1
$begingroup$
I really don't understand what you added about the even order. I also don't understand what that product there has anything to do...
$endgroup$
– DonAntonio
Nov 24 '18 at 13:52
$begingroup$
Sorry, that really didn't help at all to my understanding. Perhaps someone else...
$endgroup$
– DonAntonio
Nov 24 '18 at 14:06
$begingroup$
@DonAntonio Sorry, my thought was wrong and misleading.
$endgroup$
– Andrews
Nov 24 '18 at 14:12
add a comment |
$begingroup$
Given $p$ an odd prime, $Xinmathbb{Z}^{ntimes n}$ a matrix.
If $I_{n}+pXin operatorname{SL}(n,mathbb Z)$ has finite order, prove $X=0$.
My idea:
let $m=p^rh$, $(p,h)=1$, s.t. $(I+pX)^m=I$.
let $(I+pX)^{p^r}=I+pY$, $I=(I+pX)^m=(I+pY)^h=(I+hpY+p^2YZ)$. $Y(hI+pZ)=0$.
If we can prove $Y=0$, then the following would be quite easy.
So how to prove $hI+pZ$ is invertible? Thanks for your time and help!
abstract-algebra group-theory
asked Nov 24 '18 at 13:06
AndrewsAndrews
3831317
$endgroup$
Given $p$ an odd prime, $Xinmathbb{Z}^{ntimes n}$ a matrix.
If $I_{n}+pXin operatorname{SL}(n,mathbb Z)$ has finite order, prove $X=0$.
My idea:
let $m=p^rh$, $(p,h)=1$, s.t. $(I+pX)^m=I$.
let $(I+pX)^{p^r}=I+pY$, $I=(I+pX)^m=(I+pY)^h=(I+hpY+p^2YZ)$. $Y(hI+pZ)=0$.
If we can prove $Y=0$, then the following would be quite easy.
So how to prove $hI+pZ$ is invertible? Thanks for your time and help!
abstract-algebra group-theory
abstract-algebra group-theory
asked Nov 24 '18 at 13:06
AndrewsAndrews
3831317
asked Nov 24 '18 at 13:06
AndrewsAndrews
3831317
edited Nov 26 '18 at 19:26
Andrews
asked Nov 24 '18 at 13:06
AndrewsAndrews
3831317
asked Nov 24 '18 at 13:06
AndrewsAndrews
3831317
asked Nov 24 '18 at 13:06
AndrewsAndrews
3831317
3831317
2
$begingroup$
Have you already done anything? Did you try some examples, or some cases of $;p;$ ? Anything at all?
$endgroup$
– DonAntonio
Nov 24 '18 at 13:12
1
$begingroup$
I really don't understand what you added about the even order. I also don't understand what that product there has anything to do...
$endgroup$
– DonAntonio
Nov 24 '18 at 13:52
$begingroup$
Sorry, that really didn't help at all to my understanding. Perhaps someone else...
$endgroup$
– DonAntonio
Nov 24 '18 at 14:06
$begingroup$
@DonAntonio Sorry, my thought was wrong and misleading.
$endgroup$
– Andrews
Nov 24 '18 at 14:12
add a comment |
2
$begingroup$
Have you already done anything? Did you try some examples, or some cases of $;p;$ ? Anything at all?
$endgroup$
– DonAntonio
Nov 24 '18 at 13:12
1
$begingroup$
I really don't understand what you added about the even order. I also don't understand what that product there has anything to do...
$endgroup$
– DonAntonio
Nov 24 '18 at 13:52
$begingroup$
Sorry, that really didn't help at all to my understanding. Perhaps someone else...
$endgroup$
– DonAntonio
Nov 24 '18 at 14:06
$begingroup$
@DonAntonio Sorry, my thought was wrong and misleading.
$endgroup$
– Andrews
Nov 24 '18 at 14:12
2
2
$begingroup$
Have you already done anything? Did you try some examples, or some cases of $;p;$ ? Anything at all?
$endgroup$
– DonAntonio
Nov 24 '18 at 13:12
$begingroup$
Have you already done anything? Did you try some examples, or some cases of $;p;$ ? Anything at all?
$endgroup$
– DonAntonio
Nov 24 '18 at 13:12
1
1
$begingroup$
I really don't understand what you added about the even order. I also don't understand what that product there has anything to do...
$endgroup$
– DonAntonio
Nov 24 '18 at 13:52
$begingroup$
I really don't understand what you added about the even order. I also don't understand what that product there has anything to do...
$endgroup$
– DonAntonio
Nov 24 '18 at 13:52
$begingroup$
Sorry, that really didn't help at all to my understanding. Perhaps someone else...
$endgroup$
– DonAntonio
Nov 24 '18 at 14:06
$begingroup$
Sorry, that really didn't help at all to my understanding. Perhaps someone else...
$endgroup$
– DonAntonio
Nov 24 '18 at 14:06
$begingroup$
@DonAntonio Sorry, my thought was wrong and misleading.
$endgroup$
– Andrews
Nov 24 '18 at 14:12
$begingroup$
@DonAntonio Sorry, my thought was wrong and misleading.
$endgroup$
– Andrews
Nov 24 '18 at 14:12
add a comment |
1 Answer
1
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$begingroup$
Assume the claim is false. Let $m>1$ be minimal with the property that $Xne0$ with $1+pX$ having order $m$ exists.
If $m=ab$ is composite with $a,b>1$, note that
$$(I+pX)^a=I+sum_{j=1}^a{achoose j}p^{j}X^j =I+pX'$$
with $X':=sum_{j=1}^a{achoose j}p^{j-1}X^jinBbb Z^{ntimes n}$. Then from $(I+pX')^b=(I+pX)^m=I$ and the minimality of $m$, we conclude that $X'=0$. But then $(1+pX)^a=I$, also contradicting minimality of $m$.
We conclude that the minimla $m$ cannot be composite. As ist certainly cannot be $=1$ either, we conclude that $m$ is prime.
If $(I+pX)^m=I$ with $Xne 0$, let $rge 1$ be maximal such that we can write $X=p^{r-1}Y$ with $YinBbb Z^{ntimes n}$.
Then
$$tag1 I=(I+pX)^m=(I+p^rY)^m=I+mp^rY+{mchoose 2}p^{2r}Y^2+p^{3r}cdot (ldots)$$
implies that $mp^rY$ must be a multiple of $p^{2r}$. This means that $m$ is a multiple of $p^r$, i.e., $m=p$ and $r=1$.
As $p$ is odd, the number $mchoose 2$ in $(1)$ is a multiple of $p$ so that
$$I=(I+pY)p=I+p^2Y+p^3cdot(ldots) $$
and so $Y$ is a multiple of $p$, contradiction.
answered Nov 24 '18 at 14:16
Hagen von EitzenHagen von Eitzen
277k22269496
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Assume the claim is false. Let $m>1$ be minimal with the property that $Xne0$ with $1+pX$ having order $m$ exists.
If $m=ab$ is composite with $a,b>1$, note that
$$(I+pX)^a=I+sum_{j=1}^a{achoose j}p^{j}X^j =I+pX'$$
with $X':=sum_{j=1}^a{achoose j}p^{j-1}X^jinBbb Z^{ntimes n}$. Then from $(I+pX')^b=(I+pX)^m=I$ and the minimality of $m$, we conclude that $X'=0$. But then $(1+pX)^a=I$, also contradicting minimality of $m$.
We conclude that the minimla $m$ cannot be composite. As ist certainly cannot be $=1$ either, we conclude that $m$ is prime.
If $(I+pX)^m=I$ with $Xne 0$, let $rge 1$ be maximal such that we can write $X=p^{r-1}Y$ with $YinBbb Z^{ntimes n}$.
Then
$$tag1 I=(I+pX)^m=(I+p^rY)^m=I+mp^rY+{mchoose 2}p^{2r}Y^2+p^{3r}cdot (ldots)$$
implies that $mp^rY$ must be a multiple of $p^{2r}$. This means that $m$ is a multiple of $p^r$, i.e., $m=p$ and $r=1$.
As $p$ is odd, the number $mchoose 2$ in $(1)$ is a multiple of $p$ so that
$$I=(I+pY)p=I+p^2Y+p^3cdot(ldots) $$
and so $Y$ is a multiple of $p$, contradiction.
answered Nov 24 '18 at 14:16
Hagen von EitzenHagen von Eitzen
277k22269496
$endgroup$
add a comment |
$begingroup$
Assume the claim is false. Let $m>1$ be minimal with the property that $Xne0$ with $1+pX$ having order $m$ exists.
If $m=ab$ is composite with $a,b>1$, note that
$$(I+pX)^a=I+sum_{j=1}^a{achoose j}p^{j}X^j =I+pX'$$
with $X':=sum_{j=1}^a{achoose j}p^{j-1}X^jinBbb Z^{ntimes n}$. Then from $(I+pX')^b=(I+pX)^m=I$ and the minimality of $m$, we conclude that $X'=0$. But then $(1+pX)^a=I$, also contradicting minimality of $m$.
We conclude that the minimla $m$ cannot be composite. As ist certainly cannot be $=1$ either, we conclude that $m$ is prime.
If $(I+pX)^m=I$ with $Xne 0$, let $rge 1$ be maximal such that we can write $X=p^{r-1}Y$ with $YinBbb Z^{ntimes n}$.
Then
$$tag1 I=(I+pX)^m=(I+p^rY)^m=I+mp^rY+{mchoose 2}p^{2r}Y^2+p^{3r}cdot (ldots)$$
implies that $mp^rY$ must be a multiple of $p^{2r}$. This means that $m$ is a multiple of $p^r$, i.e., $m=p$ and $r=1$.
As $p$ is odd, the number $mchoose 2$ in $(1)$ is a multiple of $p$ so that
$$I=(I+pY)p=I+p^2Y+p^3cdot(ldots) $$
and so $Y$ is a multiple of $p$, contradiction.
answered Nov 24 '18 at 14:16
Hagen von EitzenHagen von Eitzen
277k22269496
$endgroup$
add a comment |
$begingroup$
Assume the claim is false. Let $m>1$ be minimal with the property that $Xne0$ with $1+pX$ having order $m$ exists.
If $m=ab$ is composite with $a,b>1$, note that
$$(I+pX)^a=I+sum_{j=1}^a{achoose j}p^{j}X^j =I+pX'$$
with $X':=sum_{j=1}^a{achoose j}p^{j-1}X^jinBbb Z^{ntimes n}$. Then from $(I+pX')^b=(I+pX)^m=I$ and the minimality of $m$, we conclude that $X'=0$. But then $(1+pX)^a=I$, also contradicting minimality of $m$.
We conclude that the minimla $m$ cannot be composite. As ist certainly cannot be $=1$ either, we conclude that $m$ is prime.
If $(I+pX)^m=I$ with $Xne 0$, let $rge 1$ be maximal such that we can write $X=p^{r-1}Y$ with $YinBbb Z^{ntimes n}$.
Then
$$tag1 I=(I+pX)^m=(I+p^rY)^m=I+mp^rY+{mchoose 2}p^{2r}Y^2+p^{3r}cdot (ldots)$$
implies that $mp^rY$ must be a multiple of $p^{2r}$. This means that $m$ is a multiple of $p^r$, i.e., $m=p$ and $r=1$.
As $p$ is odd, the number $mchoose 2$ in $(1)$ is a multiple of $p$ so that
$$I=(I+pY)p=I+p^2Y+p^3cdot(ldots) $$
and so $Y$ is a multiple of $p$, contradiction.
answered Nov 24 '18 at 14:16
Hagen von EitzenHagen von Eitzen
277k22269496
$endgroup$
Assume the claim is false. Let $m>1$ be minimal with the property that $Xne0$ with $1+pX$ having order $m$ exists.
If $m=ab$ is composite with $a,b>1$, note that
$$(I+pX)^a=I+sum_{j=1}^a{achoose j}p^{j}X^j =I+pX'$$
with $X':=sum_{j=1}^a{achoose j}p^{j-1}X^jinBbb Z^{ntimes n}$. Then from $(I+pX')^b=(I+pX)^m=I$ and the minimality of $m$, we conclude that $X'=0$. But then $(1+pX)^a=I$, also contradicting minimality of $m$.
We conclude that the minimla $m$ cannot be composite. As ist certainly cannot be $=1$ either, we conclude that $m$ is prime.
If $(I+pX)^m=I$ with $Xne 0$, let $rge 1$ be maximal such that we can write $X=p^{r-1}Y$ with $YinBbb Z^{ntimes n}$.
Then
$$tag1 I=(I+pX)^m=(I+p^rY)^m=I+mp^rY+{mchoose 2}p^{2r}Y^2+p^{3r}cdot (ldots)$$
implies that $mp^rY$ must be a multiple of $p^{2r}$. This means that $m$ is a multiple of $p^r$, i.e., $m=p$ and $r=1$.
As $p$ is odd, the number $mchoose 2$ in $(1)$ is a multiple of $p$ so that
$$I=(I+pY)p=I+p^2Y+p^3cdot(ldots) $$
and so $Y$ is a multiple of $p$, contradiction.
answered Nov 24 '18 at 14:16
Hagen von EitzenHagen von Eitzen
277k22269496
answered Nov 24 '18 at 14:16
Hagen von EitzenHagen von Eitzen
277k22269496
answered Nov 24 '18 at 14:16
Hagen von EitzenHagen von Eitzen
277k22269496
answered Nov 24 '18 at 14:16
Hagen von EitzenHagen von Eitzen
277k22269496
277k22269496
add a comment |
add a comment |
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$begingroup$
Have you already done anything? Did you try some examples, or some cases of $;p;$ ? Anything at all?
$endgroup$
– DonAntonio
Nov 24 '18 at 13:12
1
$begingroup$
I really don't understand what you added about the even order. I also don't understand what that product there has anything to do...
$endgroup$
– DonAntonio
Nov 24 '18 at 13:52
$begingroup$
Sorry, that really didn't help at all to my understanding. Perhaps someone else...
$endgroup$
– DonAntonio
Nov 24 '18 at 14:06
$begingroup$
@DonAntonio Sorry, my thought was wrong and misleading.
$endgroup$
– Andrews
Nov 24 '18 at 14:12